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Let the total number of marbles be x No of gold-plated marbles, = 3, No of diamond marbles, = 4 No of silver marble = x - 3 - 4 = (x - 7) Pr(both were silver) = \(\frac{(x - 7)}{x}\) x \(\frac{(x - 8)}{(x - 1)}\) ( since selection is without replacement will affect the second pick) = \(\frac{(x - 7)}{x}\) x \(\frac{(x - 8)}{(x - 1)}\) = \(\frac{1}{15}\) \(\frac{(x^2 - 15x + 56 )}{(x^2 - x) }\) = \(\frac{1}{15}\), cross multiply = 15x\(^2\) - 225x + 840 = x\(^2\) - x = 14x\(^2\) - 224x + 840 = 0 Divide through by 14 = x\(^2\) - 16x + 60 = 0 = x\(^2\) - 10x - 6x + 60 = 0 = x(x - 10) - 6(x - 10) = 0 = (x - 10)(x - 6) = 0 = x = 6 or 10 x = 10 ( total marble should not be less than the sum of both gold and diamond marbles) Recall, Silver marbles = x - 7 = 10 - 7 = 3 THEREFORE, the number of silver marbles = 3.

The term of the given G.P. are (x + 2), (4x + 3), and (7x + 24) (a) Common ratio = \(\frac{(4x + 3)}{(x + 2)}\) = \(\frac{(7x + 24)}{(4x + 3)}\) cross multiply (4x + 3)(4x + 3) = (x + 2)(7x + 24) 16x\(^2\) + 24x + 9 = 7x\(^2\) + 38x + 48 16x\(^2\) - 7x\(^2\) + 24x - 38x + 9 - 48 = 0 9x\(^2\) - 14x - 39 = 0 9x\(^2\) - 27x + 13x - 39 = 0 (from 9x\(^2\) x 39) 9x(x - 3) + 13(x -3) = 0 (9x + 13)(x - 3) = 0 (9x + 13) = 0 or (x - 3) = 0 x = \(\frac{-13}{9}\) or 3 (b) When x = \(\frac{-13}{9}\) Common ratio = \(\frac{(4x + 3)}{(x + 2)}\) = \(\frac{(4(\frac{-13}{9}) + 3)}{((\frac{-13}{9}) + 2)}\) = \(\frac{\frac{-25}{9}}{\frac{5}{9}}\) = \(\frac{- 25}{9}\) x \(\frac{9}{5}\) = - 5. Common ratio,r = - 5 When x = 3, Common ratio = \(\frac{(4(3) + 3)}{((3) + 2)}\) = \(\frac{15}{5}\) = 3. Common ratio, r = 3

(a) Age Frequency 5 2 6 2x - 1 7 y + 2 8 4 9 2 10 y - 1 \(\sum\)f = 20 2 + 2x - 1 + y + 2 + 4 +2 + y - 1 = 20 2x + 2y + 8 = 20 2x + 2y = 20 - 8 = 12 divide through by 2 x + y = 6 - - - - - - - - - - - - - -(i) Also, \(\frac{\text{x}}{\text{y}}\) = \(\frac{1}{2}\) y = 2x - - - - - - - - - - - - (ii) put y = 2x into eqn -- - - -(i) x + y = 6 = x + 2x = 6 3x = 6 x = 2 put x = 2 into eqn - - - - - (ii) y = 2x = 2(2) = 4 (b) AGE Frequency F Fx 5 2 2 10 6 2(2) - 1 = 3 3 18 7 4 + 2 = 6 6 42 8 4 4 32 9 2 2 18 10 4 - 1 = 3 3 30 \(\sum\) f = 20, \(\sum\)fx = 150 Mean,\(\overline{x}\) = \(\frac{\sum fx}{\sum f}\) = \(\frac{150}{20}\) = 7.5 THEREFORE, mean age = 7.5 years.

(a) The diagram above is the illustration of the given information. (b) Let |OB| = x, and |OC| = h From \(\triangle\)OBC Tan 48º = \(\frac{h}{x}\) h = x tan48 - - - - - - - - - - - - -(i) From \(\triangle\)AOC tan 40º = \(\frac{h}{46 + x}\) h = tan 40(46 + x) = 46(tan40) + x(tan40) - - - - - - - - - - -(ii) Equate -- - - -eqn (i) and (ii) x tan 48 = 40(tan46) + x(tan40) x tan 48 - x tan 40 = 40 tan 46 x(tan 48 - tan 40) = 40 tan 46 x(1.1106 - 0.8391) = 38.5986 x(0.2715) = 38.5986 x = \(\frac{38.5986}{0.2715}\) = 142.1680 Height of pole = from eqn (i) = 142.1680 x 1.1106 = 157.8918 ≈ 157.9 m ( to one decimal place)

Given diameter d = 26cm radius r = \(\frac{26}{2}\) = 13cm \(\overline{OQ}\) = radius NOTE: \(\overline{OR}\) bisects \(\overline{PQ}\) and \(\triangle\)POQ, \(\overline{PR}\) = \(\overline{RQ}\) Cos(Q\(\widehat{O}\)R) = \(\frac{5}{13}\) (Q\(\widehat{O}\)R) = cos\(^{-1}\)((\(\frac{5}{13}\)) = 67.38º P\(\widehat{O}\)R = 2 x Q\(\widehat{O}\)R = 2 x 67.38 = 134.76 ≈ 135( nearest whole number) (b) Area of minor segment = Area of minor sector P\(\widehat{O}\)Q - Area of triangle P\(\widehat{O}\)Q = [\(\frac{P\widehat{O}Q}{360}\) x \(\pi\) x |OQ|\(^2\)] - [\(\frac{1}{2}\) x |OQ|\(^2\) x sin(P\(\widehat{O}\)Q)] = [\(\frac{135}{360}\) x \(\frac{22}{7}\) x 13\(^2\)] - [\(\frac{1}{2}\) x 13\(^2\) x Sin(134.76)] = 198.745 - 60 = 138.745 = 139cm\(^2\) (to the nearest whole number)

let the total Acres = x Wife's Share = \(\frac{2}{5} \times x = \frac{2x}{5}\) Remaining part = x - \(\frac{2x}{5}\) = \(\frac{3x}{5}\) Part for family = \(\frac{1}{3}\) x \(\frac{3x}{5}\) = \(\frac{x}{5}\) Remaining part = x - (\(\frac{2x}{5}\) + \(\frac{x}{5}\)) = \(\frac{2x}{5}\) Total ratio = 2 + 5 + 2 = 10 Child with least share = \(\frac{2}{10}\) x \(\frac{2x}{5}\) = 8 x = \(\frac{8 \times 25}{2}\) = 100 acres. (ii) Acres wife received = \(\frac{2}{5}\) x 100 = 40 acres. (b) If $400 is deposited, the remaining amount to be paid will be = $1600 - $400 = $1200. so, P = $1200, T = 12months = 1 year, R = 25% Therefore, S.I = \(\frac{1200 \times 25 \times 1}{100}\) = $300 But, Total cost = initial deposit + principal + interest = $400 + $1,200 + $300 = $1,900

(a) Solving the eqns of the two lines simultaneously: x + 2y = 7 - - -- - - - - - -(i) x - y = 4 - - - - - - - - - - - -(ii) subtract eqn (ii) from (i) x - x + 2y - (- y) = 7 - 4 3y = 3 y = 1 put y = 1 into eqn (ii) x - 1 = 4 x = 4 + 1 = 5 The point of intersection of lines is (5, 1) ⇒ If the line passes through Origin (0, 0) and (5, 1) Using: \(\frac{y_2 - y_1}{x_2 - x_1} = \frac{y - y_1}{x - x_1}\) \(\frac{1 - 0}{5 - 0} = \frac{y - 0}{x - 0}\) \(\frac{1}{5} = \frac{\text{y}}{\text{x}}\) Therefore, the equation of the line = x - 5y = 0 (b) Given that the ratio of the interior angle to the exterior angle of a regular polygon = 4: 1 Thus, I = 4E But, the sum of the interior angles and the exterior angles of a regular polygon = 180 i.e. I + E = 180 4E + E = 180 5E = 180 E = \(\frac{180}{5}\) = 36º Since, I = 4E = 4 x 36 = 144º (ii) value of the exterior angle = 36º (i) Number of sides of the polygon = \(\frac{360}{\text{n}}\) = E, n = \(\frac{360}{36}\) = 10 (since E = 36º) (iii) Sum of the interior angles of the polygon. = (n - 2)180 = (10 - 2)180 = 8 x 180 = 1,440.

(i) 2 x T\(\widehat{S}\)Q = P\(\widehat{O}\)Q (2 times angle at circumference = angle at centre of circle) T\(\widehat{S}\)Q = \(\frac{P\widehat{O}Q}{2}\) = \(\frac{56}{28}\) = 28º T\(\widehat{Q}\)P = T\(\widehat{S}\)Q (angles in alternate segment} (ii) S\(\widehat{T}\)O + T\(\widehat{S}\)Q + S\(\widehat{V}\)P = 180º (sum of angles in a \(\triangle\)) S\(\widehat{T}\)O + 28º + 109º = 180º S\(\widehat{T}\)O = 180º - 28º - 109º = 43º Now, O\(\widehat{T}\)Q = O\(\widehat{Q}\)T (base angles of an isosceles triangle) O\(\widehat{T}\)Q + O\(\widehat{Q}\)T + P\(\widehat{O}\)Q = 180º 2(O\(\widehat{T}\)Q) = 180 - 56 = 124º O\(\widehat{T}\)Q = \(\frac{124}{2}\) = 62º Q\(\widehat{T}\)S = S\(\widehat{T}\)O + O\(\widehat{T}\)Q = 42 + 62 = 105º (b) \(\frac{2n^2 - 3n - 2 }{2n^2 + 3n + 1} \times \frac{n^2 - 1}{n^2 - 4}\) = \(\frac{(2n + 1)(n - 2)}{(2n + 1)(n + 1)} \times \frac{(n + 1)(n - 1)}{(n + 2)(n - 2)}\) After canceling common factors = \(\frac{n - 1}{n + 2}\)

(a) \(\frac{B\widehat{T}Q}{2}\) = B\(\widehat{T}\)C = C\(\widehat{T}\)Q (\(\overline{TC}\) bisects \(\angle\)BTQ) B\(\widehat{T}\)Q = B\(\widehat{A}\)T = 44º(angles in alternate segments) B\(\widehat{T}\)C = \(\frac{∠BTQ}{2}\) = \(\frac{44}{2}\) = 22º A\(\widehat{B}\)T = P\(\widehat{T}\)A = 60º (angles in alternate segments) Now, A\(\widehat{C}\)T + B\(\widehat{T}\)C = A\(\widehat{B}\)T (sum of opp. interior angles is equal to the exterior angle of a triangle) A\(\widehat{C}\)T + 22º = 60º A\(\widehat{C}\)T = 60 - 22 = 38º (b) (i) Given: Circumference of base, 2\(\pi\)r = 11 -- - - - - - - - - - -(i) Also, height, h = 6r + 3 - - - - - - - - - -(ii) From, eqn (i), 2\(\pi\)r = 11, r = \(\frac{11}{2\pi}\) = 1.75 m (ii) From equation (ii); height, h = 6r + 3 = 6(1.75) + 3 = 13.5 m (iii) Volume of cylinder, V = \(\pi r^2\)h = \(\frac{22}{7} \times 1.75^2\) x 13.5 = 130 m\(^3\)